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suppose that (X,d) is a compact metric space and $f:X\to X$ is a continuous function. Define $X_1 = f(X), X_2 = f(X_1),...,X_{i+1} = f(X_i),...$ and let $A = \bigcap_{i=1}^\infty X_i$. Is $A \subseteq F(A)$?

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What is the need for compactness in your question? – MathOverview Nov 16 '12 at 16:48

Yes. In fact, slightly more general statement is true.

Let $X, Y$ be metric spaces and $f:X\rightarrow Y$ be continuous. Let $(K_n)$ be a decreasing sequence of non-empty compact subsets of $X$. Then, $$ f(\cap_{n\in\mathbb{N}}K_n)=\cap_{n\in\mathbb{N}}f(K_n). $$ Proof follows from the nested set theorem as follows.

Let us set $$ C=\cap_{n\in\mathbb{N}}K_n. $$ Evidently. $$ f(C)\subseteq \cap_{n\in\mathbb{N}}f(K_n). $$ Conversely, let $y\in \cap_{n\in\mathbb{N}}f(K_n)$. Then, $(f^{-1}({y})\cap K_n)_{n\in\mathbb{N}}$ is decreasing sequence of non-empty compact sets. Therefore, $$ \emptyset\neq\cap_{n\in\mathbb{N}}f^{-1}({y})\cap K_n=f^{-1}({y})\cap C, $$ which proves the result.

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Yes, it is true that $A\subseteq f(A)$.

Let $a\in A$, so that $a\in X_i$ for every $i$. Since $X_i = f^i(X)$, for each $i$ choose an element $x_i$ of $X$ such that $f^i(x_i)=a$. Now consider the sequence $f(x_2), f^2(x_3), f^3(x_4), \ldots$, and let $x$ be the limit of one of its convergent subsequences (which exist, because $X$ is compact). Since $f^i(x_{i+1})$ is in $X_n$ for every $i\geq n$, we must have $x\in X_n$ for each $n$ because $X_n$ is closed (being the image of a compact set in a Hausdorff space). Therefore $x\in A$. On the other hand, $f(x)$ is the limit of a convergent subsequence of $\{f^{i+1}(x_{i+1})\}$, which is constantly $a$. Therefore $f(x)=a$ and $x\in A$, so $a\in f(A)$.

On the other hand, it's true without any assumptions on $X$ and $f$ that $f(A)\subseteq A$, so in fact we have $f(A)=A$.

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